The demand for electricity is continuously increasing. The main source of energy that provides electricity is fossil fuels. Economies like China have had an increased requirement for fossil fuels. As a result OPEC is reluctant to stop increasing oil prices. The United States Department of Energy predicts that coal will continue to dominate as the source of electrical energy well beyond 2030. Individuals, government, and industry are looking for small scale renewable energy technologies.
It costs on average between $8.00 USD and $10.00 USD to install one watt of solar cells. Wind power and other technologies have a comparable price tag. This is well out of reach for the average budget. It is still cheaper to use fossil fuels to produce electricity.
The following describes the theoretical basis for the present invention:
All objects emit blackbody radiation. The energy flux of an ideal blackbody radiator is defined by the Stephan-Boltzman law and is equal to,f=sT4 
Where “f” is the energy flux, “σ” is the Stephan-Boltzmann constant and T is the absolute temperature in Kelvin. The radiated power is equal to,P=AsT4 
“P” is the power in Watts per meter, and “A” is the surface area of the radiating surface. Radiation that strikes an object may pass right through, be absorbed, or reflected. No object completely absorbs radiation. Therefore no object is a perfect black body radiator. The emissivity “e” is the ratio of energy radiated by a particular material to energy radiated by a black body at the same temperature. Emissivity is defined as follows.e=actual radiation/blackbody radiation
The power radiated from a non-blackbody radiator is then,P=eAsT4 
Consider a long rod free floating in a uniform field of blackbody radiation. Assume that most of the rod is black except for the tip being white. Both ends of the rod are exposed to the same radiation flux f and must radiate this flux away to maintain a constant temperature. Sof=εwhitesT4white 
Andf=εblacksT4black 
The temperature white and the temperature black are then related this way.ewhitesT4white=f=εblacksT4black 
This relationship then simplifies toewhiteT4white=eblackT4black 
The temperature of the white portion is thenTwhite=(eblack/ewhite)1/4Tblack 
This equation shows that if “eblack” is not the same value as ewhite, then the temperature of the black portion is not the same as the temperature of the white portion.
It is interesting to note that the surface with the low emissivity (white) will have a higher temperature than the surface that has the high emissivity (black). The black surface will, however, heat up and cool down faster than the white surface.
As the heat is conducted down the length of the rod, from the hot side to the cooler side, some portion is lost by radiation from the surface. A long rod will have different temperatures at its ends. So, an object in thermal dynamic equilibrium with its surroundings can have two temperatures.
This has been experimentally confirmed, in a constant temperature room; with the wooden apparatus shown in figure one.
In FIG. 1 a wooden base supports a wooden vertical rod. A second wooden rod is placed horizontally on top of the vertical rod. The whole apparatus is painted black. The end of the horizontal rid is then painted white. A dual temperature digital thermometer attached to the apparatus at the two opposite ends of the horizontal rod show that the white end is consistently hotter.
Radiation emitted from a first surface, with an area “A1”, that reaches any second surface, with an area “A2”, of a two surface enclosure, is equal to the radiation emitted from the second surface that reaches the first surface. In mathematical form, this becomes,F1-2A1=F2-1A2 
If the shape of the first surface is either flat or convex, then the shape factor F1-2=1. In this situation the shape factor F2-1 yields,F2-1=A1/A2 
Consider a blackbody cavity and the opening to the cavity as a two surface enclosure where the opening to the cavity is the first surface and the cavity is the second surface. The magnitude of the radiation emitted from the first surface is equal to,e1A1sT4 
The portion of this radiation that is absorbed by the cavity, that is the second surface, is then,e2e1A1sT4 
The portion of the radiation that is reflected from the second surface is then,(1−e2)e1A1sT4 
Of this reflected radiation, the portion that reaches the opening, that is the first surface, using the shape factor F2-1 is then,(A1/A2)(1−e2)e1A1sT4 
The portion of the reflected radiation that re-contacts the second surface is,(1−A1/A2)(1−e2)e1A1sT4 
The radiation that is reflected from the second surface and is reflected from the first surface is,(1−e1)(1−e2)(A1/A2)e1A1sT4 
Each time the radiation is reflected, a portion of the initially radiated power is lost. Adding the infinite sum of all the reflections will yield the total power radiated from the first surface to the second surface. The sum of the infinite geometric series of reflections is as follows.
      P    12    =            e      1        ⁢          A      1        ⁢    s    ⁢                  ⁢                  T        1        4            ·              1                  1          -                      [                                                            (                                      1                    -                                                                  A                        1                                            /                                              A                        2                                                                              )                                ⁢                                  (                                      1                    -                                          e                      2                                                        )                                            +                                                (                                      1                    -                                          e                      1                                                        )                                ⁢                                  (                                      1                    -                                          e                      2                                                        )                                ⁢                                  (                                                            A                      1                                        /                                          A                      2                                                        )                                                      ]                              
Solving this equation, and dividing by A1 yields.
      f    12    =                    e        2            ⁢      s      ⁢                          ⁢              T        1        4                                      e          1                ⁢                              A            1                                A            2                              +              e        2            -                        e          1                ⁢                  e          2                ⁢                              A            1                                A            2                              
F2-1 can be found in a similar manner. F2-1 is equal to,
      f    21    =                    e        1            ⁢      s      ⁢                          ⁢              T        2            ⁢      4                                e          1                ⁢                              A            1                                A            2                              +              e        2            -                        e          1                ⁢                  e          2                ⁢                              A            1                                A            2                              
The Zeroth law states that any object in thermal equilibrium, must emit the same amount of radiation as it receives, or else it would heat up or cool down. To fulfill the Zeroth law, the radiation that the cavity receives from the environment, must be equal to the cavity radiation to the environment. In mathematical terms, this is equal to the following.
                    e        2            ⁢      s      ⁢                          ⁢              T        1        4                                      e          1                ⁢                              A            1                                A            2                              +              e        2            -                        e          1                ⁢                  e          2                ⁢                              A            1                                A            2                                =                    e        1            ⁢      s      ⁢                          ⁢              T        2        4                                      e          1                ⁢                              A            1                                A            2                              +              e        2            -                        e          1                ⁢                  e          2                ⁢                              A            1                                A            2                              
Simplifying the equation yields the following.T24e1=T14e2 
Solving for T2 yields.T2=T1·(e2/e1)1/4 
With an infinite number of reflections of the radiation inside the cavity, the cavity emissivity approaches 1. Therefore T2 becomes.T2=T1·(1/e1)1/4 
Letψ=(1/ε1)1/4 
T2 then becomesT2=T1ψ
More importantlyT2≠T1 
This blackbody effect can, and has, been confirmed by covering one of the two probes of a dual temperature digital thermometer with a black plastic pen lid. The covered probe will be hotter than the uncovered probe.
Seebeck Effect
A voltage can be produced in a loop consisting of two dissimilar metal wires when the two junctions are maintained at different temperatures. This voltage may be increased by connecting many junctions together in series. This is called a thermopile. Each combination of metals, or alloys, will produce a specific voltage for a specific temperature difference. There are a few well characterized combinations, such as a “K” type.
Power
The current produced by a thermocouple is related to the resistance in the wires, as stated by Ohm's law.V=IR 
Given a thermo voltage produced, the required resistance can be calculated for a specific power value as follows.P=V2/R 
When designing a thermopile, the resistance of the wires may be predicted. The resistance of the wire is calculated as follows.R=LP/A 
The resistance “R” is related with the length of the wire “L”, the cross sectional area “A”, and the constant of the material P.